c++ template argument declaration

Question

I am reading reference about enable_if on cppreference. In the example section. I have a question about the template declaration in the code:

// #4, enabled via a template parameter
template<class T,
         typename std::enable_if<
             !std::is_trivially_destructible<T>{} &&
             (std::is_class<T>{} || std::is_union<T>{}),
            int>::type = 0>
void destroy(T* t)
{
    std::cout << "destroying non-trivially destructible T\n";
    t->~T();
}

If the condition of the std::enable_if() is satisfied. The declaration can be rewrote as:

 template<class T, typename std::enable_if_t<true,int> = 0> // #1

Since the possible implementation of enable_if is:

template<class T>
struct enable_if<true, T> { typedef T type; };

The #1 is equivalent to the following statement:

template<typename T, typename int = 0>

which is an ill-formed statement. The compiler spits an error message:

error: two or more data types in declaration of 'parameter'
 template<typename T, typename int = 0>

My question is: What is the correctly way to understand the typename std::enable_if_t<true,int> = 0 in the template declaration.

---

And a similar question I have is the example followed by #4
The #5 :

// #5, enabled via a template parameter
template<class T,
    typename = std::enable_if_t<std::is_array<T>::value> >
void destroy(T* t) // note, function signature is unmodified
{
    for(std::size_t i = 0; i < std::extent<T>::value; ++i) {
        destroy((*t)[i]);
    }
}

When the condition of the enable_if() is satisfied, the declaration should be something like：

template<class T, typename = X > // X is a type name

How comes the typename=X can be compiled.

Thanks,

Rong